Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <unordered_map>
#include <queue>

int min[100000] = {0};
int min2[100000] = { 0 };
long long dp[100000] = {1, 0};
long long dp2[100000] = { 1, 0 };

int main() {
	int n, m, s, t;
	int len;
	std::cin >> n >> m >> s >> t;
	std::unordered_multimap<int, int> e;//from, to
	e.reserve(300000);
	for (int i = 0; i < m; i++) {
		int x, y;
		std::cin >> x >> y;
		e.insert(std::make_pair(x, y));
		e.insert(std::make_pair(y, x));
	}

	std::queue<std::pair<int, int>> q;
	q.push(std::make_pair(0, s));

	while (!q.empty()) {
		auto node = q.front(); q.pop();
		auto it = e.find(node.second);
		int count = e.count(node.second);
		for (int i = 0; i < count; i++, ++it) {
			auto next = node;
			next.first++;
			next.second = it->second;
			if (next.second == s || (min[next.second] > 0 && min[next.second] <= next.first)) {
				continue;
			}
			min[next.second] = next.first;
			dp[next.first]++;
			if (next.second == t) len = next.first;
			q.push(next);
		}
	}

	q.push(std::make_pair(0, t));
	while (!q.empty()) {
		auto node = q.front(); q.pop();
		auto it = e.find(node.second);
		int count = e.count(node.second);
		for (int i = 0; i < count; i++, ++it) {
			auto next = node;
			next.first++;
			next.second = it->second;
			if (next.second == t || (min2[next.second] > 0 && min2[next.second] <= next.first)) {
				continue;
			}
			min2[next.second] = next.first;
			dp2[next.first]++;
			q.push(next);
		}
	}

	long long count = 0;
	for (int i = 0; i <= len - 2; i++) {
			count += dp[i] * dp2[len - i - 2];
	}
	std::cout << count << std::endl;

	return 0;
}